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Chapter 7

Chapter 7. Continuous Distributions. Continuous random variables. Are numerical variables whose values fall within a range or interval Are measurements Can be described by density curves. Density curves. Is always on or above the horizontal axis

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Chapter 7

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  1. Chapter 7 Continuous Distributions

  2. Continuous random variables • Are numerical variables whose values fall within a range or interval • Are measurements • Can be described by density curves

  3. Density curves • Is always on or above the horizontal axis • Has an area exactly equal to one underneath it • Often describes an overall distribution • Describe what proportions of the observations fall within each range of values

  4. Can be any shape Are generic continuous distributions Probabilities are calculated by finding the area under the curve Unusual density curves

  5. How do you find the area of a triangle? P(X < 2) =

  6. What is the area of a line segment? P(X = 2) = 0 P(X < 2) = .25

  7. In continuous distributions, P(X < 2) & P(X< 2) are the same answer. Hmmmm… Is this different than discrete distributions?

  8. Shape is a trapezoid – How long are the bases? b1 = .5 b2 = .375 h = 1 P(X > 3) = P(1 < X < 3) = .5(.375+.5)(1)=.4375 .5(.125+.375)(2) =.5

  9. P(X > 1) = .75 .5(2)(.25) = .25 (2)(.25) = .5

  10. P(0.5 < X < 1.5) = .28125 .5(.25+.375)(.5) = .15625 (.5)(.25) = .125

  11. Special Continuous Distributions

  12. Uniform Distribution • Is a continuous distribution that is evenly (or uniformly) distributed • Has a density curve in the shape of a rectangle • Probabilities are calculated by finding the area under the curve How do you find the area of a rectangle? Where: a & b are the endpoints of the uniform distribution

  13. 1/.12 4.92 4.98 5.04 • The Citrus Sugar Company packs sugar in bags labeled 5 pounds. However, the packaging isn’t perfect and the actual weights are uniformly distributed with a mean of 4.98 pounds and a range of .12 pounds. • Construct the uniform distribution above. What shape does a uniform distribution have? What is the height of this rectangle? How long is thisrectangle?

  14. 1/.12 4.92 4.98 5.04 • What is the probability that a randomly selected bag will weigh more than 4.97 pounds? P(X > 4.97) = What is the length of the shaded region? .07(1/.12) = .5833

  15. 1/.12 4.92 4.98 5.04 • Find the probability that a randomly selected bag weighs between 4.93 and 5.03 pounds. What is the length of the shaded region? P(4.93<X<5.03) = .1(1/.12) = .8333

  16. 1/35 5 40 • The time it takes for students to drive to school is evenly distributed with a minimum of 5 minutes and a range of 35 minutes. • Draw the distribution What is the height of the rectangle? Where should the rectangle end?

  17. 1/35 5 40 b) What is the probability that it takes less than 20 minutes to drive to school? P(X < 20) = (15)(1/35) = .4286

  18. c) What is the mean and standard deviation of this distribution? m = (5 + 40)/2 = 22.5 s2 = (40 - 5)2/12 = 102.083 s = 10.104

  19. Normal Distributions • Symmetrical bell-shaped (unimodal) density curve • Above the horizontal axis • N(m, s) • The transition points occur at m+s • Probability is calculated by finding the area under the curve • As sincreases, the curve flattens & spreads out • As sdecreases, the curve gets taller and thinner How is this done mathematically?

  20. s s A B 6 Do these two normal curves have the same mean? If so, what is it? Which normal curve has a standard deviation of 3? Which normal curve has a standard deviation of 1? YES B A

  21. Empirical Rule • Approximately 68% of the observations fall within s of m • Approximately 95% of the observations fall within 2s of m • Approximately 99.7% of the observations fall within 3s of m

  22. 68% 71 Suppose that the height of male students at PWSH is normally distributed with a mean of 71 inches and standard deviation of 2.5 inches. What is the probability that the height of a randomly selected male student is more than 73.5 inches? 1 - .68 = .32 P(X > 73.5) = 0.16

  23. Standard Normal Density Curves Always has m = 0 & s = 1 To standardize: Must have this memorized!

  24. Strategies for finding probabilities or proportions in normal distributions • State the probability statement • Draw a picture • Calculate the z-score • Look up the probability (proportion) in the table

  25. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last less than 220 hours? Write the probability statement Draw & shade the curve P(X < 220) = .9082 Look up z-score in table Calculate z-score

  26. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last more than 220 hours? P(X>220) = 1 - .9082 = .0918

  27. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. How long must a battery last to be in the top 5%? Look up in table 0.95 to find z- score P(X > ?) = .05 .95 .05 1.645

  28. The heights of the female students at PWSH are normally distributed with a mean of 65 inches. What is the standard deviation of this distribution if 18.5% of the female students are shorter than 63 inches? What is the z-score for the 63? P(X < 63) = .185 -0.9 63

  29. Will my calculator do any of this normal stuff? • Normalpdf – use for graphing ONLY • Normalcdf – will find probability of area from lower bound to upper bound • Invnorm (inverse normal) – will find z-score for probability

  30. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last less than 220 hours? N(200,15) P(X < 220) = Normalcdf(-∞,220,200,15)=.9082

  31. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last more than 220 hours? N(200,15) P(X>220) = Normalcdf(220,∞,200,15) = .0918

  32. The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. How long must a battery last to be in the top 5%? P(X > ?) = .05 .95 .05 Invnorm(.95,200,15)=224.675

  33. The heights of female teachers at PWSH are normally distributed with mean of 65.5 inches and standard deviation of 2.25 inches. The heights of male teachers are normally distributed with mean of 70 inches and standard deviation of 2.5 inches. • Describe the distribution of differences of heights (male – female) teachers. Normal distribution with m = 4.5 & s = 3.3634

  34. 4.5 • What is the probability that a randomly selected male teacher is shorter than a randomly selected female teacher? P(X<0) = Normalcdf(-∞,0,4.5,3.3634 = .0901

  35. Ways to Assess Normality Use graphs (dotplots, boxplots, or histograms) Normal probability (quantile) plot

  36. Normal Probability (Quantile) plots The observation (x) is plotted against known normal z-scores If the points on the quantile plot lie close to a straight line, then the data is normally distributed Deviations on the quantile plot indicate nonnormal data Points far away from the plot indicate outliers Vertical stacks of points (repeated observations of the same number) is calledgranularity

  37. Why are these regions not the same width? Consider a random sample with n = 5. To find the appropriate z-scores for a sample of size 5, divide the standard normal curve into 5 equal-area regions.

  38. These would be the z-scores (from the standard normal curve) that we would use to plot our data against. Why is the median not in the “middle” of each region? Consider a random sample with n = 5. Next – find the median z-score for each region. -1.28 0 1.28 -.524 .524

  39. Normal Scores Widths of Contact Windows Suppose we have the following observations of widths of contact windows in integrated circuit chips: 3.21 2.49 2.94 4.38 4.02 3.62 3.30 2.85 3.34 3.81 Sketch a scatterplot by pairing the smallest normal score with the smallest observation from the data set & so on What should happen if our data set is normally distributed? Let’s construct a normal probability plot. The values of the normal scores depend on the sample size n. The normal scores when n = 10 are below: -1.539 -1.001 -0.656 -0.376 -0.123 0.123 0.376 0.656 1.001 1.539

  40. Notice that the boxplot is approximately symmetrical and that the normal probability plot is approximately linear. Notice that the boxplot is approximately symmetrical except for the outlier and that the normal probability plot shows the outlier. Notice that the boxplot is skewed left and that the normal probability plot shows this skewness.

  41. Are these approximately normally distributed? 50 48 54 47 51 52 46 53 52 51 48 48 54 55 57 45 53 50 47 49 50 56 53 52 What is this called? The normal probability plot is approximately linear, so these data are approximately normal. Both the histogram & boxplot are approximately symmetrical, so these data are approximately normal.

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